Homework 2 Solutions

hw02.zip

Solution Files

You can find solutions for all questions in hw02.py.

Required questions

Q1: Num eights

Write a recursive function num_eights that takes a positive integer x and returns the number of times the digit 8 appears in x. Use recursion - the tests will fail if you use any assignment statements.

def num_eights(x):
    """Returns the number of times 8 appears as a digit of x.

    >>> num_eights(3)
    0
    >>> num_eights(8)
    1
    >>> num_eights(88888888)
    8
    >>> num_eights(2638)
    1
    >>> num_eights(86380)
    2
    >>> num_eights(12345)
    0
    >>> from construct_check import check
    >>> # ban all assignment statements
    >>> check(HW_SOURCE_FILE, 'num_eights',
    ...       ['Assign', 'AugAssign'])
    True
    """
    if x % 10 == 8:
        return 1 + num_eights(x // 10)
    elif x < 10:
        return 0
    else:
        return num_eights(x // 10)

Watch the hints video below for somewhere to start:

Use Ok to test your code:

python3 ok -q num_eights

The equivalent iterative version of this problem might look something like this:

total = 0
while x > 0:
    if x % 10 == 8:
        total = total + 1
    x = x // 10
return total

The main idea is that we check each digit for a eight. The recursive solution is similar, except that you depend on the recursive call to count the occurences of eight in the rest of the number. Then, you add that to the number of eights you see in the current digit.

Q2: Ping-pong

The ping-pong sequence counts up starting from 1 and is always either counting up or counting down. At element k, the direction switches if k is a multiple of 8 or contains the digit 8. The first 30 elements of the ping-pong sequence are listed below, with direction swaps marked using brackets at the 8th, 16th, 18th, 24th, and 28th elements:

Index	1	2	3	4	5	6	7	[8]	9	10	11	12	13	14	15	[16]	17	[18]	19	20	21	22	23
PingPong Value	1	2	3	4	5	6	7	[8]	7	6	5	4	3	2	1	[0]	1	[2]	1	0	-1	-2	-3

Index (cont.)	[24]	25	26	27	[28]	29	30
PingPong Value	[-4]	-3	-2	-1	[0]	-1	-2

Implement a function pingpong that returns the nth element of the ping-pong sequence without using any assignment statements.

You may use the function num_eights, which you defined in the previous question.

Use recursion - the tests will fail if you use any assignment statements.

Hint: If you're stuck, first try implementing pingpong using assignment statements and a while statement. Then, to convert this into a recursive solution, write a helper function that has a parameter for each variable that changes values in the body of the while loop.

def pingpong(n):
    """Return the nth element of the ping-pong sequence.

    >>> pingpong(8)
    8
    >>> pingpong(10)
    6
    >>> pingpong(15)
    1
    >>> pingpong(21)
    -1
    >>> pingpong(22)
    -2
    >>> pingpong(30)
    -2
    >>> pingpong(68)
    0
    >>> pingpong(69)
    -1
    >>> pingpong(80)
    0
    >>> pingpong(81)
    1
    >>> pingpong(82)
    0
    >>> pingpong(100)
    -6
    >>> from construct_check import check
    >>> # ban assignment statements
    >>> check(HW_SOURCE_FILE, 'pingpong', ['Assign', 'AugAssign'])
    True
    """
    def helper(result, i, step):
        if i == n:
            return result
        elif i % 8 == 0 or num_eights(i) > 0:
            return helper(result - step, i + 1, -step)
        else:
            return helper(result + step, i + 1, step)
    return helper(1, 1, 1)

# Alternate solution 1
def pingpong_next(x, i, step):
    if i == n:
        return x
    return pingpong_next(x + step, i + 1, next_dir(step, i+1))

def next_dir(step, i):
    if i % 8 == 0 or num_eights(i) > 0:
        return -step
    return step

# Alternate solution 2
def pingpong_alt(n):
    if n <= 8:
        return n
    return direction(n) + pingpong_alt(n-1)

def direction(n):
    if n < 8:
        return 1
    if (n-1) % 8 == 0 or num_eights(n-1) > 0:
        return -1 * direction(n-1)
    return direction(n-1)

Watch the hints video below for somewhere to start:

Use Ok to test your code:

python3 ok -q pingpong

This is a fairly involved recursion problem, which we will first solve through iteration and then convert to a recursive solution.

Note that at any given point in the sequence, we need to keep track of the current value of the sequence (this is the value that might be output) as well as the current index of the sequence (how many items we have seen so far, not actually output).

For example, 14th element has value 0, but it's the 14th index in the sequence. We will refer to the value as x and the index as i. An iterative solution may look something like this:

def pingpong(n):
    i = 1
    x = 1
    while i < n:
        x += 1
        i += 1
    return x

Hopefully, it is clear to you that this has a big problem. This doesn't account for changes in directions at all! It will work for the first eight values of the sequence, but then fail after that. To fix this, we can add in a check for direction, and then also keep track of the current direction to make our lives a bit easier (it's possible to compute the direction from scratch at each step, see the direction function in the alternate solution).

def pingpong(n):
    i = 1
    x = 1
    is_up = True
    while i < n:
        is_up = next_dir(...)
        if is_up:
            x += 1
        else:
            x -= 1
        i += 1
    return x

All that's left to do is to write the next_dir function, which will take in the current direction and index and then tell us what direction to go in next (which could be the same direction):

def next_dir(is_up, i):
    if i % 8 == 0 or num_eights(i) > 0:
        return not is_up
    return is_up

There's a tiny optimization we can make here. Instead of calculating an increment based on the value of is_up, we can make it directly store the direction of change into the variable (next_dir is also updated, see the solution for the new version):

def pingpong(n):
    i = 1
    x = 1
    step = 1
    while i < n:
        step = next_dir(step, i)
        x += step
        i += 1
    return x

This will work, but it uses assignment. To convert it to an equivalent recursive version without assignment, make each local variable into a parameter of a new helper function, and then add an appropriate base case. Lastly, we seed the helper function with appropriate starting values by calling it with the values we had in the iterative version.

You should be able to convince yourself that the version of pingpong in the solutions has the same logic as the iterative version of pingpong above.

Video walkthrough: https://youtu.be/74gwPjgrN_k

Q3: Missing Digits

Write the recursive function missing_digits that takes a number n that is sorted in increasing order (for example, 12289 is valid but 15362 and 98764 are not). It returns the number of missing digits in n. A missing digit is a number between the first and last digit of n of a that is not in n. Use recursion - the tests will fail if you use while or for loops.

def missing_digits(n):
    """Given a number a that is in sorted, increasing order,
    return the number of missing digits in n. A missing digit is
    a number between the first and last digit of a that is not in n.
    >>> missing_digits(1248) # 3, 5, 6, 7
    4
    >>> missing_digits(1122) # No missing numbers
    0
    >>> missing_digits(123456) # No missing numbers
    0
    >>> missing_digits(3558) # 4, 6, 7
    3
    >>> missing_digits(35578) # 4, 6
    2
    >>> missing_digits(12456) # 3
    1
    >>> missing_digits(16789) # 2, 3, 4, 5
    4
    >>> missing_digits(19) # 2, 3, 4, 5, 6, 7, 8
    7
    >>> missing_digits(4) # No missing numbers between 4 and 4
    0
    >>> from construct_check import check
    >>> # ban while or for loops
    >>> check(HW_SOURCE_FILE, 'missing_digits', ['While', 'For'])
    True
    """
    if n < 10:
        return 0
    last, rest = n % 10, n // 10
    return max(last - rest % 10 - 1, 0) + missing_digits(rest)
# Alternate solution
def missing_digits_alt(n):
    def helper(n, digit):
        if n == 0:
            return 0
        last, rest = n % 10, n // 10
        if last == digit or last + 1 == digit:
            return helper(rest, last)
        return 1 + helper(n, digit - 1)
    return helper(n // 10, n % 10)

Watch the hints video below for somewhere to start:

Use Ok to test your code:

python3 ok -q missing_digits

The equivalent iterative version of this problem might look something like this:

missing = 0 while n > 10: last, rest = n % 10, n // 10 missing += max(last - rest % 10 - 1, 0) n //= 10 return missing

A tricky case for this problem was handling adjacent numbers that are the same - that's why we wrap the digit difference each recursive call with a max comparison call to 0.

Q4: Count coins

Given a positive integer total, a set of coins makes change for total if the sum of the values of the coins is total. Here we will use standard US Coin values: 1, 5, 10, 25 For example, the following sets make change for 15:

15 1-cent coins
10 1-cent, 1 5-cent coins
5 1-cent, 2 5-cent coins
5 1-cent, 1 10-cent coins
3 5-cent coins
1 5-cent, 1 10-cent coin

Thus, there are 6 ways to make change for 15. Write a recursive function count_coins that takes a positive integer total and returns the number of ways to make change for total using coins. Use the next_largest_coin function given to you to calculate the next largest coin denomination given your current coin. I.e. next_largest_coin(5) = 10.

Hint: Refer the implementation of count_partitions for an example of how to count the ways to sum up to a total with smaller parts. If you need to keep track of more than one value across recursive calls, consider writing a helper function.

def next_largest_coin(coin):
    """Return the next coin. 
    >>> next_largest_coin(1)
    5
    >>> next_largest_coin(5)
    10
    >>> next_largest_coin(10)
    25
    >>> next_largest_coin(2) # Other values return None
    """
    if coin == 1:
        return 5
    elif coin == 5:
        return 10
    elif coin == 10:
        return 25

def count_coins(total):
    """Return the number of ways to make change for total using coins of value of 1, 5, 10, 25.
    >>> count_coins(15)
    6
    >>> count_coins(10)
    4
    >>> count_coins(20)
    9
    >>> count_coins(100) # How many ways to make change for a dollar?
    242
    >>> from construct_check import check
    >>> # ban iteration
    >>> check(HW_SOURCE_FILE, 'count_coins', ['While', 'For'])                                          
    True
    """
    def constrained_count(total, smallest_coin):
        if total == 0:
            return 1
        if total < 0:
            return 0
        if smallest_coin == None:
            return 0
        without_coin = constrained_count(total, next_largest_coin(smallest_coin))
        with_coin = constrained_count(total - smallest_coin, smallest_coin)
        return without_coin + with_coin
    return constrained_count(total, 1)

Watch the hints video below for somewhere to start:

Use Ok to test your code:

python3 ok -q count_coins

This is remarkably similar to the count_partitions problem, with a few minor differences:

A maximum partition size is not given, so we need to create a helper function that takes in two arguments and also create another helper function to find the max coin.
Partition size is not linear. To get the next partition you need to call next_largest_coin

Submit

Make sure to submit this assignment by running:

python3 ok --submit

Just for Fun Questions

This question demonstrates that it's possible to write recursive functions without assigning them a name in the global frame.

Q5: Anonymous factorial

The recursive factorial function can be written as a single expression by using a conditional expression.

>>> fact = lambda n: 1 if n == 1 else mul(n, fact(sub(n, 1)))
>>> fact(5)
120

However, this implementation relies on the fact (no pun intended) that fact has a name, to which we refer in the body of fact. To write a recursive function, we have always given it a name using a def or assignment statement so that we can refer to the function within its own body. In this question, your job is to define fact recursively without giving it a name!

Write an expression that computes n factorial using only call expressions, conditional expressions, and lambda expressions (no assignment or def statements). Note in particular that you are not allowed to use make_anonymous_factorial in your return expression. The sub and mul functions from the operator module are the only built-in functions required to solve this problem:

from operator import sub, mul

def make_anonymous_factorial():
    """Return the value of an expression that computes factorial.

    >>> make_anonymous_factorial()(5)
    120
    >>> from construct_check import check
    >>> # ban any assignments or recursion
    >>> check(HW_SOURCE_FILE, 'make_anonymous_factorial', ['Assign', 'AugAssign', 'FunctionDef', 'Recursion'])
    True
    """
    return (lambda f: lambda k: f(f, k))(lambda f, k: k if k == 1 else mul(k, f(f, sub(k, 1))))
    # Alternate solution:
    #   return (lambda f: f(f))(lambda f: lambda x: 1 if x == 0 else x * f(f)(x - 1))

Use Ok to test your code:

python3 ok -q make_anonymous_factorial